Integrate the rational functions.
∫2xx2+3x+2dx.
Let 2xx2+3x+2=2xx2+2x+x+2
=2xx(x+2)+1(x+2)=2x(x+2)(x+1)
Let 2x(x+2)(x+1)=A(x+2)+B(x+1)
⇒2x(x+2)(x+1)=A(x+1)+B(x+2)(x+2)(x+1)⇒2x=Ax+A+Bx+2B→2x=x(A+B)+(A+2B).
On comparing the coefficients of x and constant term on both sides, we get
A+B=2 .....(i)
and A+2B=0 .....(ii)
On subtracting Eq. (ii)form Eq. (i), we get -B =2 ⇒ B =-2
On putting the value of B in Eq. (i),we get A-2=2 ⇒ A =4
∴∫2x(x+2)(x+1)dx=∫A(x+2)dx+∫B(x+1)dx=∫4(x+2)dx+∫(−2)(x+1)dx=4log|x+2|−2log|x+1|+C